It can be... but the resolution of the center frequency of two DDS
frequencies would be limited to half DDS LSB... so not enough. I will
try to monitor FSELECT and also to extract some SPI data if I found some
free time today or tomorrow :)
Regards,
Javier, EA1CRB
El 28/01/2012 15:35, Azelio Boriani escribió:
And I think it depends on the two frequencies loaded too. The FSELECT
selects between two phase accumulator steps. Maybe the word sent to the Rb
is manipulated to obtain two symmetric values to load.
On Sat, Jan 28, 2012 at 3:21 PM, Javier Herrero<[email protected]>wrote:
El 27/01/2012 19:27, beale escribió:
I added a bit to the "electronics" section of the FE-5680A FAQ as below.
http://www.ko4bb.com/dokuwiki/doku.php?id=precision_timing:fe5680a_faq#electronic
(Note- until today, I had the 8 and 6 digits transposed, calling it the
fe5860a. But no one noticed :-)
The updated section is below. I measured the 20 MHz input and 5.3 MHz
output of the DDS, but I'm puzzled by how the tuning resolution (4.6 mHz)
of the DDS output is divided by such a large factor to achieve 0.18 uHz
resolution at the final 10 MHz output. Can any frequency synthesizer gurus
explain how this is done?
The frequencies inside the unit are quite similar to those found in the
FRS-C, 60 and 5.3125MHz. The FRS-C excites the cavity at 60MHz x 114 -
5.3125MHz = 6.8346875GHz (a bit over the 6.834682608GHz Rb natural
resonance - so I suppose that the resonance is driven to 6.8346875GHz using
the C-Field), so I understand that the FE-5680A operates in the same way.
Since in the multiplication process the 60MHz frequency is multiplied by
114 and the 5.3125MHz only by one, 1Hz offset in the 5.3125MHz frequency
will need 1/114Hz offset in the 60MHz signal to obtain the same resonant
frequency.
I've checked that the DDS is driven by 10MHz, not 20MHz (I've just checked
it), so the 5.3125MHz is probably an image and not a fundamental DDS
output. Hence, the minimum DDS step is 2.23mHz. A change of 2.23mHz in the
5.3125MHz frequency is compensated by an approximately 20.45uHz change at
the 60MHz frequency, and so, a 3.41uHz at the 10MHz output, i.e. one part
in 3.41^-13. This lets to a factor of 19 between the adjustment attainable
directly by modifiying a 1LSB and the claimed 1.7854^-14 adjustment.
Probably this is done by modifying the duty cycle of the FSELECT signal. I
suspect that the 416.6666667 signal at FSELECT is used to produce the
modulation on the cavity excitation to perform a synchronous detection (the
same way it is done in the FRS-C at 127Hz), to obtain a null, so the null
can be slightly "moved" by variying the duty cycle at FSELECT.
I will try to play a bit more this evening :)
Best regards,
Javier
--
------------------------------------------------------------------------
Javier Herrero
Chief Technology Officer EMAIL: [email protected]
HV Sistemas S.L. PHONE: +34 949 336 806
Los Charcones, 17 FAX: +34 949 336 792
19170 El Casar - Guadalajara - Spain WEB: http://www.hvsistemas.com
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